In the secondary treatment unit of a WWTP, two activated sludge aeration tanks are operated in
parallel. Each tank has the following dimensions: 7.0 m wide by 30.0 m long by 4.3 m effective
liquid depth. The plant operating parameters are as follows:
Flow = 0.0796 m3/s
Soluble BOD5 after primary settling = 130 mg/L
MLVSS = 1,500 mg/L
MLSS = 1.40 x (MLVSS)
Settled sludge volume after 30 min = 230.0 mL/L
3Using the following assumptions, determine (a) the solids retention time, (b) the cell wastage
flow rate, and (c) the return sludge flow rate for the WWTP. Assume:
Allowable BOD in effluent = 25.0 mg/L
Suspended solids in effluent = 25.0 mg/L (30% of SS account for BOD)
Yield coefficient (Y) = 0.60 mg VSS/mg BOD5 removed
Decay rate of microorganisms (kd) = 0.060 d-